How do you differentiate #f(x)=xsqrt(abs(x-1)# using the product rule?

Answer 1

Please see the explanation section below.

We must first analyze the function because it involves the absolute value function.

Note that #abs(x-1) = {(x-1," if ",x-1 >= 0),(1-x," if ",x-1<0):}#.

Thus, we have

#f(x) = {(xsqrt(x-1)," if ",x > 1),(xsqrt(1-x)," if ",x < 1):}#.
Now we can differentiate using #d/dx (sqrtu) = 1/(2sqrtu) (du)/dx#.
#d/dx(xsqrt(x-1)) = (1)sqrt(x-1)+x 1/(2sqrt(x-1))*1#
# = (2x-2+x)/(2sqrt(x-1)) = (3x-2)/(2sqrt(x-1))#

and

#d/dx(xsqrt(1-x)) = (1)sqrt(1-x)+x 1/(2sqrt(1-x))*(-1)#
# = (2-2x-x)/(2sqrt(1-x)) = (2-3x)/(2sqrt(1-x))#.

Combining these results in

#f'(x) = {((3x-2)/(2sqrt(x-1))," if ",x > 1),((2-3x)/(2sqrt(1-x))," if ",x < 1) :}#.

Extra remarks

We can write the derivative of #absx# as #{(1," if " x > 0),(-1," if "x < 0):}#
Or, we can use #absx/x# (or its reciprocal).

This notation gives us

#d/dx (sqrtabs(x-1)) = 1/(2sqrt abs(x-1))* d/dx(abs(x-1))#
# = 1/(2sqrt(abs(x-1))) abs(x-1)/(x-1)#.
We can write #f'(x)# as
#f'(x) = (3x-2)/(2sqrt abs(x-1)) abs(x-1)/(x-1)#.
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Answer 2

To differentiate the function ( f(x) = x \sqrt{\left| x - 1 \right|} ) using the product rule, we treat it as the product of two functions: ( u(x) = x ) and ( v(x) = \sqrt{\left| x - 1 \right|} ). Then, we apply the product rule ( (uv)' = u'v + uv' ), where ( u' ) and ( v' ) represent the derivatives of ( u ) and ( v ) with respect to ( x ), respectively.

First, we find ( u'(x) ), the derivative of ( u(x) ) with respect to ( x ), which is simply ( 1 ) since the derivative of ( x ) with respect to ( x ) is ( 1 ).

Next, we find ( v'(x) ), the derivative of ( v(x) ) with respect to ( x ). To do this, we consider two cases:

  1. When ( x > 1 ): In this case, ( v(x) = \sqrt{x - 1} ). The derivative ( v'(x) ) can be found using the chain rule, yielding ( \frac{1}{2 \sqrt{x - 1}} ).

  2. When ( x < 1 ): Here, ( v(x) = \sqrt{1 - x} ). Again, using the chain rule, the derivative ( v'(x) ) is ( -\frac{1}{2 \sqrt{1 - x}} ).

Now, we apply the product rule: [ (uv)' = u'v + uv' ] [ = (1)(\sqrt{\left| x - 1 \right|}) + (x)\left( \frac{1}{2 \sqrt{x - 1}} \right) ] [ = \sqrt{\left| x - 1 \right|} + \frac{x}{2 \sqrt{x - 1}} ]

Thus, the derivative of ( f(x) = x \sqrt{\left| x - 1 \right|} ) using the product rule is: [ f'(x) = \sqrt{\left| x - 1 \right|} + \frac{x}{2 \sqrt{x - 1}} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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